The equation of a circle,which passes through the centre of the circle $x^2+y^2+8x+10y-7=0$ and is concentric with the circle $2x^2+2y^2-8x-12y-9=0$,is

If $P(\alpha, \beta)$ is the radical centre of the circles $S \equiv x^2+y^2+4x+7=0$,$S^{\prime} \equiv 2x^2+2y^2+3x+5y+9=0$ and $S^{\prime \prime} \equiv x^2+y^2+y=0$,then the length of the tangent drawn from $P$ to $S^{\prime}=0$ is

If $P$ is the point of contact of the circles $x^2+y^2+4x+4y-10=0$ and $x^2+y^2-6x-6y+10=0$,and $Q$ is their external centre of similitude,then the equation of the circle with $P$ and $Q$ as the extremities of its diameter is

The radical axis of circles $x^2+y^2+5x+4y-5=0$ and $x^2+y^2-3x+5y-6=0$ is

If the circles $x^2+y^2-6x-8y-12=0$ and $x^2+y^2-4x+6y+k=0$ are orthogonal to each other,then the value of $k$ is:

If $\left(0, \frac{3}{4}\right)$ is the radical centre of the circles $S \equiv x^2+y^2+\alpha x+6y=0$,$S^{\prime} \equiv x^2+y^2+2\alpha x+\alpha y+6=0$ and $S^{\prime\prime} \equiv x^2+y^2+6\alpha x-\alpha y+3=0$,then the distance between the radical centre and the centre of the circle $S^{\prime}=0$ is: